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\(\Rightarrow4^x=4^4:4^3=4^1\\ \Rightarrow x=1\)

ban viet lai de bai duoc ko minh ko hieu neu viet so mu thi an shift roi an so 7 la mu vi du 3^2

đề :    CMR :  1+4+4^2+4^3+..........+4^58+4^59 chia hết cho 21,85

mk chỉ viết đề thôi chứ bn thông cảm mk chưa hok dạng này

Ta thấy :

\(\dfrac{1}{4^2}>\dfrac{1}{4.5}\)

\(\dfrac{1}{5^2}>\dfrac{1}{5.6}\)

..............

\(\dfrac{1}{99^2}>\dfrac{1}{99.100}\)

\(\Rightarrow\) \(K>\dfrac{1}{4.5}+\dfrac{1}{5.6}+.....+\dfrac{1}{99.100}\)

Ta có công thức \(\dfrac{a}{b.c}=\dfrac{a}{c-b}.\left(\dfrac{1}{b}-\dfrac{1}{c}\right)\)

Dựa vào công thức ta có :

\(\dfrac{1}{4.5}=\dfrac{1}{4}-\dfrac{1}{5}\)

\(\dfrac{1}{5.6}=\dfrac{1}{5}-\dfrac{1}{6}\)

.......................

\(\dfrac{1}{99.100}=\dfrac{1}{99}-\dfrac{1}{100}\)

\(\Rightarrow\) \(K>\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+......+\dfrac{1}{99}-\dfrac{1}{100}\)

\(\Leftrightarrow\) \(K>\dfrac{1}{4}-\dfrac{1}{100}\)

\(\Rightarrow K>\dfrac{6}{25}>\dfrac{1}{5}\Rightarrow dpcm\) (1)

Ta có :

\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)

\(\dfrac{1}{5^2}< \dfrac{1}{4.5}\)

................

\(\dfrac{1}{99^2}< \dfrac{1}{98.99}\)

Dựa vào công thức \(\dfrac{a}{b.c}=\dfrac{a}{c-b}.\left(\dfrac{1}{b}-\dfrac{1}{c}\right)\) ta có :

\(K< \dfrac{1}{3.4}+\dfrac{1}{4.5}+......+\dfrac{1}{98.99}\)

\(\Rightarrow\) \(K< \dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+.......+\dfrac{1}{98}-\dfrac{1}{99}\)

\(\Rightarrow\) \(K< \dfrac{1}{3}-\dfrac{1}{99}\)

Vậy \(K< \dfrac{32}{99}< \dfrac{1}{3}\Rightarrow dpcm\) (2)

(1) ; (2) \(\Rightarrow\) \(\dfrac{1}{5}< K< \dfrac{1}{3}\)

Ai thấy đúng thì ủng hộ nha !!!

Công thức tổng quát: \(\dfrac{1}{n\left(n+1\right)}< \dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right)n}\)

=>\(\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}< K< \dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{98.99}\)

=>\(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}< K< \dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{98}-\dfrac{1}{99}\)

=>\(\dfrac{1}{4}-\dfrac{1}{100}< K< \dfrac{1}{3}-\dfrac{1}{100}\)

=>\(\dfrac{1}{4}< K< \dfrac{1}{3}\)

=>\(\dfrac{1}{5}< K< \dfrac{1}{3}\left(do\dfrac{1}{4}>\dfrac{1}{5}\right)\)

1; 5.2² + (\(x\) + 3) = 5²

   5.4  +  (\(x\) + 3) = 25

    20 + (\(x\) + 3) = 25

             \(x\) + 3 = 25 - 20

             \(x+3\) = 5

             \(x\)       = 5  - 3

            \(x\)        = 2

            Vậy \(x=2\)

2; 2³ + (\(x\) - 3²) = 5³ - 4³

   8 +  (\(x\) - 9)    = 125 - 64

   8 + (\(x\) - 9) = 61

         \(x\) - 9 = 61 - 8

         \(x\) - 9 = 53

        \(x\)        = 53  + 9

        \(x\)       = 62

        Vậy \(x\) = 62

a) 5.2² + (x + 3) = 5²

5.4 + x + 3 = 25

20 + x + 3 = 25

x + 23 = 25

x = 25 - 23

x = 2

b) 2³ + (x - 3²) = 5³ - 4³

8 + (x - 9) = 125 - 64

8 + x - 9 = 61

x - 1 = 61

x = 61 + 1

x = 62

c) 4.(x - 5) - 2³ = 2⁴.3

4x - 20 - 8 = 16.3

4x - 28 = 48

4x = 48 + 28

4x = 76

x = 76 : 4

x = 19

d) 5.(x + 7) - 10 = 2³.5

5x + 35 - 10 = 8.5

5x + 25 = 40

5x = 40 - 25

5x = 15

x = 15 : 5

x = 3

e) 7² - 7.(13 - x) = 14

49 - 91 + 7x = 14

7x - 42 = 14

7x = 14 + 42

7x = 56

x = 56 : 7

x = 8

a) \(5\cdot2^2+\left(x+3\right)=5^2\)

\(\Rightarrow x+3=5^2-5\cdot2^2\)

\(\Rightarrow x+3=25-5\cdot4\)

\(\Rightarrow x+3=5\)

\(\Rightarrow x=5-3\)

\(\Rightarrow x=2\)

b) \(2^3+\left(x-3^2\right)=5^3-4^3\)

\(\Rightarrow8+\left(x-9\right)=125-64\)

\(\Rightarrow8+x-9=61\)

\(\Rightarrow x-1=61\)

\(\Rightarrow x=61+1\)

\(\Rightarrow x=62\)

c) \(4\left(x-5\right)-2^3=2^4\cdot3\)

\(\Rightarrow4\left(x-5\right)=2^4\cdot3+2^3\)

\(\Rightarrow4\cdot\left(x-5\right)=16\cdot3+8\)

\(\Rightarrow4\cdot\left(x-5\right)=56\)

\(\Rightarrow x-5=56:4\)

\(\Rightarrow x-5=14\)

\(\Rightarrow x=19\)

d) \(5\left(x+7\right)-10=2^3\cdot5\)

\(\Rightarrow5\left(x+7\right)=8\cdot5+10\)

\(\Rightarrow5\left(x+7\right)=40+10\)

\(\Rightarrow5\left(x+7\right)=50\)

\(\Rightarrow x+7=10\)

\(\Rightarrow x=10-7\)

\(\Rightarrow x=3\)

e) \(7^2-7\left(13-x\right)=14\)

\(\Rightarrow7\left(13-x\right)=7^2-14\)

\(\Rightarrow7\left(13-x\right)=49-14\)

\(\Rightarrow7\left(13-x\right)=35\)

\(\Rightarrow13-x=5\)

\(\Rightarrow x=13-5\)

\(\Rightarrow x=8\)

f) \(5x-5^2=10\)

\(\Rightarrow5x=10+5^2\)

\(\Rightarrow5x=10+25\)

\(\Rightarrow5x=35\)

\(\Rightarrow x=\dfrac{35}{5}\)

\(\Rightarrow x=7\)

g) \(9x-2\cdot3^2=3^4\)

\(\Rightarrow9x=3^4+2\cdot3^2\)

\(\Rightarrow9x=81+2\cdot9\)

\(\Rightarrow9x=99\)

\(\Rightarrow x=\dfrac{99}{9}\)

\(\Rightarrow x=11\)

h) \(10x+2^2\cdot5=10^2\)

\(\Rightarrow10x=10^2-2^2\cdot5\)

\(\Rightarrow10x=100-4\cdot5\)

\(\Rightarrow10x=80\)

\(\Rightarrow x=\dfrac{80}{10}\)

\(\Rightarrow x=8\)

i) \(125-5\left(4+x\right)=15\)

\(\Rightarrow5\left(4+x\right)=125-5\)

\(\Rightarrow5\left(4+x\right)=120\)

\(\Rightarrow4+x=\dfrac{120}{5}\)

\(\Rightarrow4+x=24\)

\(\Rightarrow x=24-4\)

\(\Rightarrow x=20\)

j) \(2^6+\left(5+x\right)=3^4\)

\(\Rightarrow5+x=3^4-2^6\)

\(\Rightarrow5+x=81-64\)

\(\Rightarrow5+x=17\)

\(\Rightarrow x=17-5\)

\(\Rightarrow x=12\)

a: =16-2+91=14+91=105

b: =9*5+8*10-27=45+53=98

c: =32+65-3*8=8+65=73

d; \(=5^3-10^2=125-100=25\)

e: \(=4^2-3^2+1=8\)

f: =9*16-16*8-8+16*4

=16(9-8+4)-8

=16*5-8

=72

a) \(2^4-50:25+13\cdot7\)

\(=2^4-2+91\)

\(=16-2+91\)

\(=14+91\)

\(=105\)

b) \(3^2\cdot5+2^3\cdot10-3^4:3\)

\(=9\cdot5+8\cdot10-3^3\)

\(=45+80-27\)

\(=98\)

c) \(2^5+5\cdot13-3\cdot2^3\)

\(=32+65-3\cdot8\)

\(=32+65-24\)

\(=73\)

d) \(5^{13}:5^{10}-5^2\cdot2^2\)

\(=5^{13-10}-\left(5\cdot2\right)^2\)

\(=5^3-10^2\)

\(=125-100\)

\(=25\)

e) \(4^5:4^3-3^9:3^7+5^0\)

\(=4^{5-3}-3^{9-7}+1\)

\(=4^2-3^2+1\)

\(=16-9+1\)

\(=8\)

f) \(3^2\cdot2^4-2^3\cdot4^2-2^3\cdot5^0+4^2\cdot2^2\)

\(=3^2\cdot4^2-2^3\cdot4^2-2^3\cdot1+4^2\cdot2^2\)

\(=4^2\cdot\left(3^2-2^3+2^2\right)-2^3\)

\(=4^2\cdot\left(9-8+4\right)-8\)

\(=16\cdot5-8\)

\(=72\)

1: 2*2*2*2*2*2=2^6

2: x*x*x*x=x^4

3: =7^4*6^4=42^4

4: =5^2*3^2*4^2=60^2

5: =2*3^2*5^5

6: =10^3*10^3=10^6